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OLEGan [10]
3 years ago
8

While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the

Physics
1 answer:
Vladimir [108]3 years ago
4 0
24-15=9 m/s slower in 12 seconds. So 9/12 m/s² slower. Therefore the acceleration is -0,75 m/s²
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A 2.40 kg ball is attached to an unknown spring and allowed to oscillate. The figure below shows a graph of the ball’s position
irga5000 [103]

(a) The period of the oscillation is 0.8 s.

(b) The frequency of the oscillation is 1.25 Hz.

(c) The angular frequency of the oscillation is 7.885 rad/s.

(d) The amplitude of the oscillation is 3 cm.

(e) The force constant of the spring is 148.1 N/m.

The given parameters:

  • <em>Mass of the ball, m = 2.4 kg</em>

<em />

From the given graph, we can determine the missing parameters.

The amplitude of the wave is the maximum displacement, A = 3 cm

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T = 0.8 s

The frequency of the oscillation is calculated as follows;

f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz

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Learn more about general wave equation here: brainly.com/question/25699025

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3 years ago
Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was
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Answer:

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Explanation:

Givens  

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f_{y}=n-mg=0\\ n=mg\\f_{x}=F-f_{s,max} =0\\ f_{s,max}=F=ma\\

Substituting (3) into (1), we get:

f_{s,max}= μ_s*m*g

Equating this equation with (4), we get:

ma=  μ_s*m*g

 μ_s=a/g

      =4.18

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